Let G be a simple connected planar graph with at least three vertices. Prove that if every face of G has degree 3, then G has an even number of faces

Let G be a simple connected planar graph with at least three vertices. Prove that if every face of G has degree 3, then G has an even number of faces

I've looked around and haven't found any questions similar to this.

I'm trying to prove this. I know that every face shares two edges so if I were to use Euler's formula 3f = 2m. Im not really sure where to go from there.

• You should use Euler formula, and the fact that $3f$ gives you twice the number of edges because each edge is counted twice. From there, you simply need to argue that the number of vertices is an integer. – Fabio Somenzi May 22 '18 at 8:27
• So would it be something like since 3f = 2m, m = 3/2f so f has to be even? – Akshay Kumar May 22 '18 at 8:32
• Exactly. Otherwise, plugging that value for $m$ in Euler's formula, you find that $n-2$ is not an integer. – Fabio Somenzi May 22 '18 at 8:34
• I get what your saying, does this mean that n - 3/2f + f = 2. which only true for even faces – Akshay Kumar May 22 '18 at 8:36
• Yes, that's the proof. – Fabio Somenzi May 22 '18 at 12:41

I know that every face shares two edges so if I were to use Euler's formula $3f = 2m$
What's $m$? What does Euler's formula have to do with it? In fact, being planar is largely irrelevant: the same is true on surfaces of arbitrary genus.
If every face has degree $3$, every face has $3$ edges; but every edge has two faces, so $E = \frac{3}{2}F$ where $E$ is the number of edges and $F$ is the number of faces; then $2E = 3F$ and by the fundamental theorem of arithmetic $2$ must be a factor either of $3$ (which it isn't) or of $F$.